Release less energy than it uses
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
Answer:
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Answer : The half life of 28-Mg in hours is, 6.94
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:

where,
k = rate constant
t = time passed by the sample = 48.0 hr
a = initial amount of the reactant disintegrate = 53500
a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520
Now put all the given values in above equation, we get


Now we have to calculate the half-life.



Therefore, the half life of 28-Mg in hours is, 6.94
Answer:
hey you wanna get it right try this one: 48.0 kcal
was released... at constant pressure.
Explanation: